An electric circuit switches instantaneously from a 5 volt battery to a 18 volt battery 3 seconds after being turned on. Sketch on a sheet of paper a graph the battery voltage against time. Then fill in the formulas below for the function represented by your graph.
for t < V(t) =
for t ≥ V(t) =
At what point or points is your function dicontinuous?
t =
doesn't it have a step disconunity at t=3? it goes from 5 to 18 in a step function.https://www.khanacademy.org/math/differential-equations/laplace-transform/properties-of-laplace-transform/v/laplace-transform-of-the-unit-step-function
I don't think we need to use the Laplace transform, since they are specifying V(t) as a piecewise function.
Just specify two values for V to the left and right of t=3
To sketch the graph of the battery voltage against time, we can consider the two separate intervals: before the switch (t < 3 seconds) and after the switch (t ≥ 3 seconds).
Before the switch (t < 3 seconds), the battery voltage remains constant at 5 volts. So, the graph will be a horizontal line at y = 5, for all values of t less than 3 seconds.
After the switch (t ≥ 3 seconds), the battery voltage changes to 18 volts. So, the graph will be a horizontal line at y = 18, for all values of t greater than or equal to 3 seconds.
We can now fill in the formulas for the function represented by the graph:
For t < 3, V(t) = 5 volts.
For t ≥ 3, V(t) = 18 volts.
The function is discontinuous at t = 3 seconds because there is an immediate change in the battery voltage from 5 volts to 18 volts at that time.